Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 10;
const int N = 100000;
const int logN = 18;
const int MAX = 10;
int n, m;
vector<int> A[N];
int lca[N][logN];
int lcaLive[N][logN][MAX] = {};
int Live[N][MAX] = {};
int living[N] = {};
int d[N];
void merge(int* merd, int a[MAX], int b[MAX]) {
  int p1 = 0;
  int p2 = 0;
  int cnt = 0;
  while (cnt < MAX) {
    if (a[p1] == b[p2]) {
      merd[cnt] = a[p1];
      p1++;
      p2++;
      cnt++;
    } else if (a[p1] < b[p2]) {
      merd[cnt] = a[p1];
      p1++;
      cnt++;
    } else {
      merd[cnt] = b[p2];
      p2++;
      cnt++;
    }
  }
}
void dfs(int u, int p, int h) {
  d[u] = h;
  lca[u][0] = p;
  for (int i = 1; i < logN; ++i) lca[u][i] = lca[lca[u][i - 1]][i - 1];
  merge(lcaLive[u][0], Live[u], Live[p]);
  for (int i = 1; i < logN; ++i)
    merge(lcaLive[u][i], lcaLive[u][i - 1], lcaLive[lca[u][i - 1]][i - 1]);
  for (auto v : A[u]) {
    if (v != p) dfs(v, u, h + 1);
  }
}
int LCA(int u, int v) {
  int a = u, b = v;
  if (d[a] > d[b]) swap(a, b);
  int dif = d[b] - d[a];
  for (int i = 0; dif > 0; i++) {
    if (dif & 1) b = lca[b][i];
    dif >>= 1;
  }
  if (a == b) return a;
  for (int i = logN - 1; i >= 0; --i) {
    if (lca[a][i] != lca[b][i]) {
      a = lca[a][i];
      b = lca[b][i];
    }
  }
  return lca[a][0];
}
void getLive(int* live, int u, int up) {
  int a = u;
  int dif = up;
  for (int i = 0; i < MAX; ++i) live[i] = inf;
  for (int i = 0; dif > 0; i++) {
    if (dif & 1) {
      int h[MAX];
      for (int i = 0; i < MAX; ++i) h[i] = live[i];
      merge(live, h, lcaLive[a][i]);
      a = lca[a][i];
    }
    dif >>= 1;
  }
}
int main() {
  int q;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    A[u].push_back(v);
    A[v].push_back(u);
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < MAX; ++j) Live[i][j] = inf;
  for (int i = 0; i < m; ++i) {
    int c;
    cin >> c;
    c--;
    if (living[c] < MAX) Live[c][living[c]++] = i;
  }
  dfs(0, 0, 0);
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    u--;
    v--;
    int x1[MAX], x2[MAX], ans[MAX];
    if (u == v) {
      for (int i = 0; i < MAX; ++i) ans[i] = Live[u][i];
    } else {
      int anc = LCA(u, v);
      getLive(x1, u, d[u] - d[anc]);
      getLive(x2, v, d[v] - d[anc]);
      merge(ans, x1, x2);
    }
    int k = 0;
    for (int i = a - 1; i >= 0; --i)
      if (ans[i] != inf) {
        k = i + 1;
        break;
      }
    cout << k << ' ';
    for (int i = 0; i < k; ++i) cout << ans[i] + 1 << ' ';
    cout << '\n';
  }
}